If M is the product of all positive integers greater than 59 and less

Question

If M is the product of all positive integers greater than 59 and less than 71, then what is the greatest integer n for which  \frac{M}{6^n}  is an integer?

A. 5
B. 7
C. 9
D. 11
E. 13

Bunuel · Accepted Answer

We need to find the power of of 6 in 60*61*62*63*64*65*66*67*68*69*70.

6 = 2*3.

Since the power of 2 is higher in 60*61*62*63*64*65*66*67*68*69*70, than the power of 3, then we'll have as many 6's as there are 3's. So, basically we need to find the power of 3 in 60*61*62*63*64*65*66*67*68*69*70:

One 3 in 60;
Two 3's in 63 (63 = 3^2*7);
One 3 in 66;
One 3 in 69.

Total of five 3's.

Therefore the power of 3, as well as power of 6 in 60*61*62*63*64*65*66*67*68*69*70 is 5.

Answer: A.

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Hope it helps.

Kinshook · Answer

Asked: If M is the product of all positive integers greater than 59 and less than 71, then what is the greatest integer n for which  \frac{M}{6^n}  is an integer?

M = 70!/59!

Greatest power of 3 in 70! = 23 + 7 + 2 = 32
Greatest power of 3 in 59! = 19 + 6 + 2 = 27
Greatest power of 3 in M = 70!/59! = 5
Greatest power of 6 in M = 5

IMO A

Nidzo · Answer

What we have is 70! - 59! and this is being divided by 6^n

We take the greatest prime in 6 which 3 and see how many times it goes into each of the factorials using the formula  n/p + n/p^2 + n/p^3....+ n/p^k  where n is the factorial we are dealing with and p is the largest prime (3)

70/3=23 , 70/9=7 , 70/27=2 
 total = 32

59/3=19 , 59/9=6 , 59/27=2
 total = 27

32 - 27 =5 
 Answer A

Kinshook · Answer

Asked: If M is the product of all positive integers greater than 59 and less than 71, then what is the greatest integer n for which  \frac{M}{6^n}  is an integer?

M = 60*61*62*63*64*65*66*67*68*69*70 = (2^2*3*5)*61*(2*31)*(3*3*7)*(2^6)*(5*13)*(2*3*11)*67*(2*2*17)*(3*23)*(2*5*7) = 2^13*3^5*Integer = 6^5*Integer
The greatest integer n for which  \frac{M}{6^n}  is an integer = 5

IMO A

lesgooo · Answer

Hi Bunuel Sir.
But wouldn't we need 2 and 3 both to be the factors of the number from 60-70 to be divisible by 6
Say what I thought is that 60 consists of a 2& a 3. Also 66. So aren't those the only 2 numbers divisible by 6. 
Please explain where I am going wrong. I believe this is a fundamental fallacy on my part.

Bunuel · Answer

The gap is that the whole product must be divisible by 6, not each individual number.

A factor of 6 needs one 2 and one 3, but these can come from different numbers in the product. For example, 62 is not divisible by 3 and 63 is not even, but 62 * 63 is divisible by 6 because 62 contributes a factor of 2 and 63 contributes factors of 3.

So we count total factors of 2 and 3 in the entire product. Since there are more than enough 2s, the number of 6s is limited by the number of 3s. From 60, 63, 66, and 69, we get 1 + 2 + 1 + 1 = 5 factors of 3, so the greatest value of n is 5.

If M is the product of all positive integers greater than 59 and less

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If M is the product of all positive integers greater than 59 and less

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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